Đáp án:a)$\frac{\sqrt{x}-1}{\sqrt{x}-2}$
b)$\frac{\sqrt{2}}{\sqrt{2}-1}$
Giải thích các bước giải:
a)M=$(2-\frac{4\sqrt{x}}{x-1}+\frac{2}{\sqrt{x}-1}):\frac{2x-4\sqrt{x}}{x-1}$
=$(\frac{2(x-1)-4\sqrt{x}+2(\sqrt{x}+1}{x-1}):\frac{2\sqrt{x}(\sqrt{x}-2)}{x-1}$
=$(\frac{2x-2-4\sqrt{x}+2\sqrt{x}+2}{x-1})·\frac{x-1}{2\sqrt{x}(\sqrt{x}-2)}$
=$\frac{2\sqrt{x}(\sqrt{x}-1)}{x-1}·\frac{x-1}{2\sqrt{x}(\sqrt{x}-2)}$
=$\frac{\sqrt{x}-1}{\sqrt{x}-2}$
b)Khi $x=3+2\sqrt{2}$
⇒M=$\frac{\sqrt{x}-1}{\sqrt{x}-2}=\frac{\sqrt{x}-2+1}{\sqrt{x}-2}$
=$1+\frac{1}{\sqrt{x}-2}$
⇒M=$1+\frac{1}{\sqrt{3+2\sqrt{2}}-2}=1+\frac{1}{\sqrt{(\sqrt{2}+1)^{2}}-2}$
=$1+\frac{1}{|\sqrt{2}+1|-2}=1+\frac{1}{\sqrt{2}+1-2}=1+\frac{1}{\sqrt{2}-1}$
=$\frac{\sqrt{2}}{\sqrt{2}-1}$